# Work done to pump water out of a conical tank into a window above it

Water is pumped from a conical tank of top radius 3 ft and a height of 5 ft to a window 10 ft above the tank. The tank is completely full of water. How much work is done?

This is what I have so far: $$\int_0^{15}\pi(\text{radius of slice})^2 \cdot (62.4) (15-h)\,dh$$ I am having a problem figuring out the radius of the slice

• Hint: chop the conical tank into infinitesimal cylinders.
– user137731
Commented Nov 10, 2015 at 19:25
• This is what i have so far: integral [0, 15] pi(radius of slice)^2 * (62.4) (15-h) I am having a problem figuring out the radius of the slice. Commented Nov 10, 2015 at 20:02
• You might tell us poor Europeans where that $62.4$ comes from. (Also you might set the upper limit of integration to $5$ instead of $15$.) Commented Nov 10, 2015 at 23:20
• @TonyK The density of water at atmospheric pressure and $\approx 50^\circ \text{F}$ is $62.4 \frac{\text{lbm}}{\text{ft}^3}$. Commented Feb 14 at 6:54
• Thanks for pounds mass, instead of the vague pounds, which could be pounds-force or pounds-mass. Commented Feb 14 at 9:07

To draw the problem in a coordinate system, choose a positive $x$ axis in the downward direction since the motion is vertical. Take the window at $x = 0$, the top of the tank at $x = 10$ and the bottom of the tank at $x = 15$. The partition of water is therefore the closed interval $[10,15]$ on the $x$ axis.

Take the radius of the cone at $x_i$, the function is $9-3x_i/5$. Then take an element of volume as a circular disk having thickness $\Delta_i x$ and the radius of the cone at $x_i$, the volume is $\Delta_i V = \pi (9-3x_i/5)^2 \Delta_i x$. Then take $62.4$ as the weight of water in pounds per cubic foot, the force required to pump an element is $\Delta_i F = 62.4 \pi (9-3x_i/5)^2 \Delta_i x$.

If $\Delta_i x$ is close to $0$, then the distance through which an element moves is approximately $x_i$. Thus the work done pumping an element to the window is $\Delta_i W = 62.4 \pi x_i (9-3x_i/5)^2 \Delta_i x$ and the integral is:

$$W = 62.4 \pi \int_{10}^{15} \! x(9-3x/5)^2 \, \textrm{d}x$$

We'll assume no frictional losses, since we aren't given any information about the flow.

Let the density of water be $$\rho = 62.4 \frac{\text{lbm}}{\text{ft}^3}$$, the base radius of the cone be $$R = 3 \text{ ft}$$, the height of the cone be $$h = 5 \text{ ft}$$, and the vertical distance from the top of the cone to the final water elevation be $$d = 10 \text{ ft}$$. Of course, the gravitational acceleration is $$g = 1 \frac{\text{lbf}}{\text{lbm}}$$.

Consider an infinitessimally thin layer of water in the tank at height $$y$$ above the ground, with a height $$d y$$. By similar triangles, the radius of this layer of water is $$\frac{y}{h} R$$. Therefore, the mass of this layer is $$d m = \rho d V = \pi \rho \frac{y^2}{h^2} R^2 d y$$. The change in gravitational potential energy in bringing this layer to the final elevation is $$d U = g (h + d - y) d m$$. Therefore, the total energy required is

$$\Delta U = \int_{y = 0}^{y = h} d U = \pi \rho g \frac{R^2}{h^2} \int_0^h y^2 (h + d - y) d y$$

$$= \pi \rho g \frac{R^2}{h^2} \left[ \frac{1}{3} (h + d) y^3 - \frac{1}{4} y^4 \right]_0^h$$

$$= \frac{1}{3} \pi \rho g R^2 h d \left[ 1 + \frac{1}{4} \frac{h}{d} \right]$$

Notice that if we let $$V = \frac{1}{3} \pi R^2 h$$ be the volume of the cone, then the mass of the water is $$m = \rho V = \frac{1}{3} \pi \rho R^2 h$$, meaning

$$\Delta U = m g d \left[ 1 + \frac{1}{4} \frac{h}{d} \right]$$

This result is actually quite intuitive, since you are bringing the entire water mass to the same final height, and the center of mass of a cone is located $$\frac{1}{4}$$ of the height from its base.

I'll leave it to you to substitute the numerical values.

• This is a nice approach but it would be better if it were better synchronized with the formula shown in the question. That is, use $h$ as the variable of integration and use some other symbol to represent the total height of the tank. Note that the integral in the question is your integral with the numbers filled in except for the unknown (to OP) radius of the slice (which was the actual question) and the incorrect upper bound of integration. Commented yesterday
• I also suggest writing $\left[d+\frac14h\right]$, which is even more intuitive than $d\left[1+\frac14\frac hd\right]$. Commented yesterday

Never work with force if you have energy:

The volume of a cylinder with slope $$r = a \ z$$ is $$V= \int_0^h \pi\ (a\ z)^2 dz =\frac{1}{3} \ \pi \ a ^2 h^3$$ Its potential energy wrt. to the base is

$$V_{pot} = \int_0^h g \rho \ z \ \ \pi (a \ z)^2 dz =\frac{g \ \rho}{4} \ \pi \ a ^2 h^4$$

Work done is

$$W = g \ \rho \ V H - V_{pot} = \frac{1}{3}\ \pi \ a \ H^3 \ \left(\frac{3\ g \ H}{4}-1\right)$$